Trigonometric Functions

  $co{s}^{3}3x+co{s}^{3}5x={\left(2cos4xcosx\right)}^3$  

$={\left(cos5x+cos3x\right)}^3$            

$co{s}^{3}3x+co{s}^{3}5x$            

$\Rightarrow cosx.cos3x.cos4x.cos5x=0$            

$\therefore x=\left(2n+1\right)\frac{\pi }{2},\left(2n+1\right)\frac{\pi }{6},\left(2n+1\right)\frac{\pi }{8},\left(2n+1\right)\frac{\pi }{10}$            

$\Rightarrow$ Smallest +ve values of x is $\frac{\pi }{10}$  i.e. 18°.

tanx + tan (x+π/3) + tan (x-π/3) = 3
tanx + (tanx+√3)/ (1-√3tanx) + (tanx-√3)/ (1+√3tanx) = 3
tanx + (8tanx)/ (1-3tan²x) = 3
(tanx (1-3tan²x) + 8tanx)/ (1-3tan²x) = 3
(3 (3tanx – tan³x)/ (1-3tan²x) = 3
⇒ 3tan3x = 3
tan3x = 1

tan α = 1/x, tan β = 2/x, tan γ = 3/x
Now α + β + γ = 180°
⇒ tan α + tan β + tan γ = tan α tan β tan γ
1/x + 2/x + 3/x = (123)/ (xxx)
⇒ x² = 1

f (x) = x√ (1 – x²)
∴ 1 - x² ≥ 0
⇒ x ∈ [-1, 1]
Put x = cos θ
⇒ f (θ) = cos θ|sin θ|
If sin θ > 0, f (θ) = sin θ cos θ
= (sin 2θ)/2, f (θ) ∈ [-1/2, 1/2]
if sin θ < 0, f () = sin cos
= - (sin 2θ)/2, f (θ) ∈ [-1/2, 1/2]

2x=tan (π/9)+tan (7π/18)
=sin (π/9+7π/18) / cos (π/9)cos (7π/18)
=sin (π/2) / cos (π/9)cos (7π/18)
=1 / cos (π/9)cos (7π/18)
=1 / cos (π/9)sin (π/2−7π/18)
=1 / cos (π/9)sin (π/9)
⇒x=1 / 2cos (π/9)sin (π/9)
=1 / sin (2π/9)=cosec (2π/9)

Again 2y=tan (π/9)+tan (5π/18)
⇒2y=sin (π/9+5π/18) / cos (π/9)cos (5π/18)
=sin (7π/18) / sin (π/2−π/9)sin (π/2−5π/18)
=sin (7π/18) / sin (7π/18)sin (4π/18) = cosec (2π/9)
⇒|x−2y|=0

α=max {2? sin³?2? cos³? }
=max {2? sin³? 2? cos³? }=2¹?
β=min {2? sin³?2? cos³? }=2? ¹?
α¹/? +β¹/? = b/8
⇒4+1/4 = b/8
⇒17/4 = b/8 ⇒ b=-34
Again α¹/? β¹/? =c/8
⇒4*1/4 = c/8
⇒c=8
⇒c−b=8+34=42

(81)^sin²x + (81)^cos²x = 30.
(81)^sin²x + (81)^ (1-sin²x) = 30.
Let y = 81^sin²x.
y + 81/y = 30
y² - 30y + 81 = 0
(y - 3) (y - 27) = 0
⇒ y = 3 or y = 27.
Either 81^sin²x = 3 ⇒ 3^ (4sin²x) = 3¹ ⇒ sin²x = 1/4 ⇒ sin x = ±1/2. x = π/6, 5π/6.
OR, 81^sin²x = 27 ⇒ 3^ (4sin²x) = 3³ ⇒ sin²x = 3/4 ⇒ sin x = ±√3/2. x = π/3, 2π/3.
(as 0 ≤ x ≤ π)
Total possible solutions = 4.

y = √ (2cos²α / (sinα cosα) + 1/sin²α)
y = √ (2cotα + cosec²α)
y = √ (2cotα + 1 + cot²α) = √ (1 + cotα)²) = |1 + cotα|.
Given α is in a range where 1+cotα is negative, y = -1 - cotα.
dy/dα = - (-cosec²α) = cosec²α.
At α = 5π/6, dy/dα = cosec² (5π/6) = (1/sin (5π/6)² = (1/ (1/2)² = 2² = 4.